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Maurer--Cartan structure equation

Statement

Assume ACω. Let θ be the left Maurer--Cartan form of a Lie group G, with values in g=TeG. Write dθ for its componentwise exterior derivative from Finite-dimensional vector-valued forms and their exterior derivative. For vector fields A,B, define the bracket-valued wedge by

[θθ](A,B)=[θ(A),θ(B)]G[θ(B),θ(A)]G=2[θ(A),θ(B)]G.

Then the normalized left Maurer--Cartan structure equation is

dθ+12[θθ]=0.

The countable-choice assumption is used exactly through the supplied Maurer--Cartan, invariant-field, and tangent-bracket results.

Facts & Assumptions

Given: ACω, a Lie group G with identity e, its left Maurer--Cartan form θ, and g=TeG with the transported left-invariant-field bracket.

[F1]

ACω is countable choice. The Axiom of Countable Choice (ACω).

[F2]

Finite-dimensional vector-valued forms and their componentwise exterior derivative are defined by scalar dual evaluation. Finite-dimensional vector-valued forms and their exterior derivative.

[F3]

Every θp is a linear isomorphism with inverse d(Lp)e, and θ(XL)=X. Maurer--Cartan form is a pointwise isomorphism and left invariant.

[F4]

The transported tangent bracket satisfies [XL,YL]=[X,Y]GL. Lie bracket on the tangent space of a Lie group.

[F5]

The tangent bracket is bilinear and alternating. The tangent space at the identity is a Lie algebra.

[F6]

Every Xg has a unique smooth left-invariant extension. Left-invariant vector fields evaluate isomorphically at the identity.

[F7]

For a scalar one-form α, dα(A,B)=A(α(B))B(α(A))α([A,B]). The exterior derivative by the invariant vector-field formula.

Proof

technique · direct
1.1

Bilinearity of the bracket [F5] and smoothness of θ [F3] show in any basis of g that the displayed bracket-wedge has smooth components; alternation follows by exchanging A and B. Thus it is a well-defined g-valued two-form. Because the bracket is alternating, [F5] also gives [θθ](A,B)=2[θ(A),θ(B)]G.

F3F5algebra
1.2

Let X,Yg and use their left-invariant extensions from [F6]. For every λg, [F2], [F7], and [F3] give λ(dθ(XL,YL))=XL(λ(θ(YL)))YL(λ(θ(XL)))λ(θ([XL,YL]))=λ([X,Y]G), because the first two functions are the constants λ(Y) and λ(X) and [F4] identifies the bracket field. Since linear functionals separate points, dθ(XL,YL)=[X,Y]G.

F2F3F4F6F7algebra
2.1

On the same fields, step 1.1 and [F3] yield 12[θθ](XL,YL)=[X,Y]G. Adding this to step 1.2 proves the structure equation on every pair of left-invariant fields.

F3step 1.1step 1.2algebra
3.1

Fix pG and U,VTpG. Put X=θp(U) and Y=θp(V). By [F3], θp(XpL)=X=θp(U) and θp(YpL)=Y=θp(V); injectivity of θp gives XpL=U and YpL=V. Step 2.1 therefore makes the structure equation vanish on the arbitrary pair (U,V) at p, proving it globally.

F3F6step 2.1
4.1

A Lie group is nonempty. If dimG=0, both two-forms are uniquely zero; if dimG=1, every alternating two-form is zero and the equation again holds. Lie groups are boundaryless by convention, and no metric, nondegeneracy, or endpoint is involved. The stated ACω is inherited through [F3], [F4], [F5], and [F6]; componentwise differentiation and the pointwise spanning argument add no choice. The theorem is one equality, not a biconditional.

F1F2F3F4F5F6F7step 1.1step 1.2step 2.1step 3.1

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